2
YES 我的答案：
#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;
int main() {
int n;
cin >> n;
switch (n) {
case 2:
case 4:
case 6:
case 7:
cout << "YES" << endl;
break;
case 1:
case 3:
case 5:
cout << "NO" << endl;
break;
default:
cout << "非法输入\n";
}
return 0;
}
1 2 3
Invalid
3 3 3
9 我的答案：
#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;
int main() {
double a,b,c,C;
cin >> a>>b>>c;
if(a+b>c && a+c>b && b+c>a)
cout<<a+b+c<<endl;
else
cout<<"Invalid"<<endl;
return 0;
}
10 4 9
7 我的答案：
#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;
int main() {
int n,x,y,p;
cin>>n>>x>>y;
p=ceil(y*1.0/x);
if(p>=n)
cout<<"0"<<endl;
else
cout<<n-p<<endl;
//cout<<n-ceil(y*1.0/x)<<endl;
return 0;
}
4 2 8
2->4->8 我的答案：
#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;
void swap(int *x, int *y) {
int temp;
temp = *y;
*y = *x;
*x = temp;
}
int main() {
int a, b, c;
cin >> a >> b >> c;
if (a > b)
swap(a, b);
if (b > c)
swap(b, c);
if (a > b)
swap(a, b);
cout << a << "->" << b << "->" << c << endl;
return 0;
} 【描述】
某电网执行“阶梯电价”，居民用户电价分为两个“阶梯”：月用电量50千瓦时（含50千瓦时）以内的，电价为0.53元/千瓦时；超过50千瓦时的，超出部分的用电量，电价上调0.05元/千瓦时。请编写程序计算电费。结果保留2位小数。
【输入】
一行中输入用电量。
【输出】
一行中输出电费。
【输入示例】
100
cost = 55.50 我的答案：
#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;
int main() {
double n, cost;
cin >> n;
if (n <= 50)
cost = n * 0.53;
else
cost = 50 * 0.53 + (n - 50) * (0.53 + 0.05);
cout << "cost = " << fixed << setprecision(2) << cost << endl;
return 0;
} 
100
10005 我的答案：
#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;
int main() {
int x, fx;
cin >> x;
if (x < 0)
fx = x + 1;
else if (x >= 0 && x < 100)
fx = x * x * x;
else if (x >= 100)
fx = x * x + 5;
cout << fx << endl;
return 0;
}
74
2.4 我的答案：
#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;
int main() {
int n;
double grade;
cin >> n;
if (n >= 60 && n <= 100) {
grade = (n - 50) / 10.0;
cout << fixed << setprecision(1) << grade << endl;
} else if (n > 0 && n < 60)
cout << "Failed" << endl;
else
cout << "Invalid" << endl;
return 0;
}
1 2 3
Don't make a triangle
4 2 5
General triangle 我的答案：
#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;
int main() {
double a, b, c;
const double EPSILON = 0.0000001;
cin >> a >> b >> c;
if (a + b > c && a + c > b && b + c > a) {
if (fabs(a - b) <= EPSILON && fabs(a - c) <= EPSILON && fabs(b - c) <= EPSILON) {
cout << "Equilateral triangle" << endl;
} else if (fabs(a - b) <= EPSILON || fabs(a - c) <= EPSILON || fabs(b - c) <= EPSILON) {
if (fabs(a - b) <= EPSILON || fabs(a - c) <= EPSILON || fabs(b - c) <= EPSILON) {
cout << "lsosceles Right triangle" << endl;
} else
cout << "lsosceles triangle" << endl;
} else if (fabs(a * a + b * b - c * c) <= EPSILON || fabs(a * a + c * c - b * b) <= EPSILON
|| fabs(c * c + b * b - a * a) <= EPSILON) {
cout << "Right triangle" << endl;
} else
cout << "General triangle" << endl;
} else
cout << "Don't make a triangle" << endl;
return 0;
} 
2.1 8.9 3.5
x1=-0.44;x2=-3.80 我的答案：
#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;
int main() {
double a, b, c, x1, x2, pd;
cin >> a >> b >> c;
pd = b * b - 4 * a * c;
if (a == 0 && b == 0 && c == 0)
cout << "Infinitely solution" << endl;
if (a == 0 && b == 0 && c != 0)
cout << "No solution" << endl;
if (a == 0 && b != 0)
cout << "x=" << fixed << setprecision(2) << -c * 1.0 / b << endl;
if (a != 0 && pd == 0)
cout << "x1=x2=" << fixed << setprecision(2) << -b * 1.0 / (2 * a) << endl;
if (a != 0 && pd > 0)
cout << "x1=" << fixed << setprecision(2) << (-b + sqrt(pd)) * 1.0 / (2 * a) << ";x2=" << fixed << setprecision(2) <<
(-b - sqrt(pd)) * 1.0 / (2 * a) << endl;
if (a != 0 && pd < 0)
cout << "Imaginary root" << endl;
return 0;
}
91
A 我的答案：
#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;
int main() {
int score;
cin >> score;
switch(score/10){
case 10:
case 9:
cout<<"A"<<endl;
break;
case 8:
cout<<"B"<<endl;
break;
case 7:
cout<<"C"<<endl;
break;
case 6:
cout<<"D"<<endl;
break;
case 5:
case 4:
case 3:
case 2:
case 1:
case 0:
cout<<"E"<<endl;
break;
default:
cout<<"非法输入"<<endl;
break;
}
return 0;
}